Suppose a positive random variable $ T $ with an absolutely continouse distribution $F$ is our model for duration of life of an individual; consider the so called "excess life distribution"

$$P(T>x_{0n} +a_n y|T>x_{0n}) = 1 - \tilde F_n(y) = \frac{1 - F(x_{0n}+ a_n y)}{ 1- F(x_{0n})} .$$

This is a very interesting object to study when we speak abut longevity: what is the probability that the individual will live longer than $x_{0n}+ a_n y$, given that this individual already lived for longer than $x_{0n}$. It is especially interesting to study when $x_{0n}$ is large.

According to the so-called Balkemaa-de Haan-Pickands theorem, if there is a sequence of constants $a_n$ such that for $x_{0n}\to\infty$ there is a limit,

$$\tilde F_n(y) \to G(y), \quad y>0,$$

then this $G$ can only be of the following form:

$$ G(y) = \begin{cases} 1- (1+ \theta y/\sigma)^{-1/\theta},  \; {\rm for } \;  \theta>0,\\ 1- e^{-y/\sigma}, \; {\rm for} \; \theta =0 .\end{cases}  $$

Now we follow what is said in the book DemgraphyBook, Lecture 15.1*:

Denote $\mu(z) = f(z)/[1-F(z)]$ is the force of mortality, corresponding to the distribution $F$. If the asymptotic relation

$$ \int _{x_{0n}}^{x_{0n}+a_n y} \mu(z) dz \sim \mu(x_{0n}) a_n y,\quad x_{0n} \to \infty, \quad (1) $$

holds, the only choice for $a_n$ is $a_n \sim c/\mu(x_{0n}),$ with $ c $ a generic notation for constant, and the only possible limit for $ \tilde F_n $ is the exponential distribution:

$$ 1- \tilde F_n (y) \to \exp [- c y]  . $$

What is unusual and needs investigation is this: the (1) is true for all distributions we discussed in the book, and yet the longevity data (for those over 90) for New Zealand population, does not quite agree with the exponential distribution it has to follow. Moreover, serious researchers studied similar data on longevity in other countries and also suggested long tail distribution above. But ... (1) is true and the limit must be exponential :-). Of course, one should understand what is happening.
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